# Palindromic primes in base 16

Palindromic primes in base 16 is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Find palindromic primes   n   in base 16,   where   n   <   50010

## 11l

```F is_prime(a)
I a == 2
R 1B
I a < 2 | a % 2 == 0
R 0B
L(i) (3 .. Int(sqrt(a))).step(2)
I a % i == 0
R 0B
R 1B

L(n) 500
V h = hex(n)
I h == reversed(h) & is_prime(n)
print(h, end' ‘ ’)
print()```
Output:
```2 3 5 7 B D 11 101 151 161 191 1B1 1C1
```

## Action!

```INCLUDE "H6:SIEVE.ACT"

BYTE FUNC Palindrome(CHAR ARRAY s)
BYTE l,r

l=1 r=s(0)
WHILE l<r
DO
IF s(l)#s(r) THEN RETURN (0) FI
l==+1 r==-1
OD
RETURN (1)

PROC IntToHex(INT i CHAR ARRAY hex)
CHAR ARRAY digits="0123456789ABCDEF"
BYTE d

hex(0)=0
WHILE i#0
DO
d=i MOD 16
hex(0)==+1
hex(hex(0))=digits(d+1)
i==/16
OD
RETURN

BYTE Func IsPalindromicPrime(INT i CHAR ARRAY hex
BYTE ARRAY primes)

BYTE d
INT rev,tmp

IF primes(i)=0 THEN
RETURN (0)
FI

IntToHex(i,hex)
IF Palindrome(hex) THEN
RETURN (1)
FI
RETURN (0)

PROC Main()
DEFINE MAX="499"
BYTE ARRAY primes(MAX+1)
INT i,count=[0]
CHAR ARRAY hex(5)

Put(125) PutE() ;clear the screen
Sieve(primes,MAX+1)
FOR i=2 TO MAX
DO
IF IsPalindromicPrime(i,hex,primes) THEN
Print(hex) Put(32)
count==+1
FI
OD
PrintF("%EThere are %I palindromic primes",count)
RETURN```
Output:
```2 3 5 7 B D 11 101 151 161 191 1B1 1C1

There are 13 palindromic primes
```

## ALGOL 68

```BEGIN  # find primes that are palendromic in base 16 #
# sieve the primes to 499 #
[]BOOL prime = PRIMESIEVE 499;
# returns an array of the digits of n in the specified base #
PRIO DIGITS = 9;
OP   DIGITS = ( INT n, INT base )[]INT:
IF   INT v := ABS n;
v < base
THEN v # single dogit #
ELSE   # multiple digits #
[ 1 : 10 ]INT result;
INT d pos := UPB result + 1;
INT v     := ABS n;
WHILE v > 0 DO
result[ d pos -:= 1 ] := v MOD base;
v OVERAB base
OD;
result[ d pos : UPB result ]
FI # DIGITS # ;
# returns TRUE if the digits in d form a palindrome, FALSE otherwise #
OP   PALINDROMIC = ( []INT d )BOOL:
BEGIN
INT  left := LWB d, right := UPB d;
BOOL is palindromic := TRUE;
WHILE left < right AND is palindromic DO
is palindromic := d[ left ] = d[ right ];
left          +:= 1;
right         -:= 1
OD;
is palindromic
END;
# print the palendromic primes in the base 16 #
STRING base digits = "0123456789ABCDEF";
FOR n TO UPB prime DO
IF prime[ n ] THEN
# have a prime #
IF []INT d = n DIGITS 16;
PALINDROMIC d
THEN
# the prime is palindromic in base 16 #
print( ( " " ) );
FOR c FROM LWB d TO UPB d DO print( ( base digits[ d[ c ] + 1 ] ) ) OD
FI
FI
OD
END```
Output:
``` 2 3 5 7 B D 11 101 151 161 191 1B1 1C1
```

## AppleScript

```on isPrime(n)
if ((n < 4) or (n is 5)) then return (n > 1)
if ((n mod 2 = 0) or (n mod 3 = 0) or (n mod 5 = 0)) then return false
repeat with i from 7 to (n ^ 0.5) div 1 by 30
if ((n mod i = 0) or (n mod (i + 4) = 0) or (n mod (i + 6) = 0) or ¬
(n mod (i + 10) = 0) or (n mod (i + 12) = 0) or (n mod (i + 16) = 0) or ¬
(n mod (i + 22) = 0) or (n mod (i + 24) = 0)) then return false
end repeat

return true
end isPrime

set digits to "0123456789ABCDEF"'s characters
set output to {"2"} -- Take "2" as read.
repeat with n from 3 to 499 by 2 -- All other primes are odd.
if (isPrime(n)) then
-- Only the number's hex digit /values/ are needed for testing.
set vals to {}
repeat until (n = 0)
set vals's beginning to n mod 16
set n to n div 16
end repeat
-- If they're palindromic, build a text representation and append this to the output.
if (vals = vals's reverse) then
set hex to digits's item ((vals's beginning) + 1)
repeat with i from 2 to (count vals)
set hex to hex & digits's item ((vals's item i) + 1)
end repeat
set output's end to hex
end if
end if
end repeat

return output

```
Output:
```{"2", "3", "5", "7", "B", "D", "11", "101", "151", "161", "191", "1B1", "1C1"}
```

## Arturo

```palindrome?: function [a][
and? -> prime? a
-> equal? digits.base:16 a reverse digits.base:16 a
]

print map select 1..500 => palindrome? 'x -> upper as.hex x
```
Output:
`2 3 5 7 B D 11 101 151 161 191 1B1 1C1`

## AWK

```# syntax: GAWK -f PALINDROMIC_PRIMES_IN_BASE_16.AWK
BEGIN {
start = 1
stop = 499
for (i=start; i<=stop; i++) {
hex = sprintf("%X",i)
if (is_prime(i) && hex == reverse(hex)) {
printf("%4s%1s",hex,++count%10?"":"\n")
}
}
printf("\nPalindromic primes %d-%d: %d\n",start,stop,count)
exit(0)
}
function is_prime(x,  i) {
if (x <= 1) {
return(0)
}
for (i=2; i<=int(sqrt(x)); i++) {
if (x % i == 0) {
return(0)
}
}
return(1)
}
function reverse(str,  i,rts) {
for (i=length(str); i>=1; i--) {
rts = rts substr(str,i,1)
}
return(rts)
}
```
Output:
```   2    3    5    7    B    D   11  101  151  161
191  1B1  1C1
Palindromic primes 1-499: 13
```

## Delphi

Works with: Delphi version 6.0

```function IsPrime(N: int64): boolean;
{Fast, optimised prime test}
var I,Stop: int64;
begin
if (N = 2) or (N=3) then Result:=true
else if (n <= 1) or ((n mod 2) = 0) or ((n mod 3) = 0) then Result:= false
else
begin
I:=5;
Stop:=Trunc(sqrt(N+0.0));
Result:=False;
while I<=Stop do
begin
if ((N mod I) = 0) or ((N mod (I + 2)) = 0) then exit;
Inc(I,6);
end;
Result:=True;
end;
end;

function IsPalindrome(N, Base: integer): boolean;
{Test if number is the same forward or backward}
var S1,S2: string;
begin
S2:=ReverseString(S1);
Result:=S1=S2;
end;

procedure ShowPalindromePrimes16(Memo: TMemo);
var I: integer;
var Cnt: integer;
var S: string;
begin
Cnt:=0;
for I:=1 to 1000-1 do
if IsPrime(I) then
if IsPalindrome(I,16) then
begin
Inc(Cnt);
S:=S+Format('%4X',[I]);
If (Cnt mod 5)=0 then S:=S+CRLF;
end;
end;
```
Output:
```   2   3   5   7   B
D  11 101 151 161
191 1B1 1C1 313 373
3B3
Count=16
Elapsed Time: 2.116 ms.

```

## EasyLang

```fastfunc isprim num .
i = 2
while i <= sqrt num
if num mod i = 0
return 0
.
i += 1
.
return 1
.
func reverse s .
while s > 0
e = e * 16 + s mod 16
s = s div 16
.
return e
.
digs\$[] = strchars "0123456789abcdef"
func\$ hex n .
if n = 0
return ""
.
return hex (n div 16) & digs\$[n mod 16 + 1]
.
for i = 2 to 499
if isprim i = 1
if reverse i = i
write hex i & " "
.
.
.
```
Output:
```2 3 5 7 b d 11 101 151 161 191 1b1 1c1
```

## F#

This task uses Extensible Prime Generator (F#)

```let rec fN g=[yield g%16; if g>15 then yield! fN(g/16)]
primes32()|>Seq.takeWhile((>)500)|>Seq.filter(fun g->let g=fN g in List.rev g=g)|>Seq.iter(printf "%0x "); printfn ""
```
Output:
```2 3 5 7 b d 11 101 151 161 191 1b1 1c1
```

## Factor

```USING: kernel math.parser math.primes prettyprint sequences
sequences.extras ;

500 primes-upto [ >hex ] [ dup reverse = ] map-filter .
```
Output:
```V{
"2"
"3"
"5"
"7"
"b"
"d"
"11"
"101"
"151"
"161"
"191"
"1b1"
"1c1"
}
```

## FreeBASIC

```Function isprime(num As Ulongint) As Boolean
For i As Integer = 2 To Sqr(num)
If (num Mod i = 0) Then Return False
Next i
Return True
End Function

Function reverse(Byval text As String) As String
Dim As String text2 = text
Dim As Integer x, lt = Len(text)
For x = 0 To lt Shr 1 - 1
Swap text2[x], text2[lt - x - 1]
Next x
Return text2
End Function

Dim As Integer inicio = 2, final = 499, cont = 0

For i As Integer = inicio To final
Dim As String hexi = Str(Hex(i))

If isprime(i) = True And hexi = reverse(hexi) Then
cont += 1
Print Hex(i); "  ";
End If
Next i

Print !"\n\nEncontrados"; cont; " primos palindrómicos entre " & inicio & " y " & final
Sleep
```
Output:
```2  3  5  7  B  D  11  101  151  161  191  1B1  1C1

Encontrados 13 primos palindrómicos entre 2 y 499```

## Go

Library: Go-rcu
```package main

import (
"fmt"
"rcu"
"strconv"
"strings"
)

func reverse(s string) string {
chars := []rune(s)
for i, j := 0, len(chars)-1; i < j; i, j = i+1, j-1 {
chars[i], chars[j] = chars[j], chars[i]
}
return string(chars)
}

func main() {
fmt.Println("Primes < 500 which are palindromic in base 16:")
primes := rcu.Primes(500)
count := 0
for _, p := range primes {
hp := strconv.FormatInt(int64(p), 16)
if hp == reverse(hp) {
fmt.Printf("%3s ", strings.ToUpper(hp))
count++
if count%5 == 0 {
fmt.Println()
}
}
}
fmt.Println("\n\nFound", count, "such primes.")
}
```
Output:
```Primes < 500 which are palindromic in base 16:
2   3   5   7   B
D  11 101 151 161
191 1B1 1C1

Found 13 such primes.
```

## J

```   palindromic16=: (-: |.)@hfd@>
hfd@> (#~ palindromic16) p: i. p:inv 500
2
3
5
7
b
d
11
101
151
161
191
1b1
1c1
```

## jq

Works with: jq

Works with gojq, the Go implementation of jq

This entry uses a generator that produces an unbounded stream of arrays of the form [dec, hex], where `dec` is the palindromic prime as a JSON number, and `hex` is the JSON string corresponding to its hexadecimal representation.

For a suitable implementation of `is_prime`, see e.g. Erdős-primes#jq.

```# '''Preliminaries'''

def emit_until(cond; stream): label \$out | stream | if cond then break \$out else . end;

# decimal number to exploded hex array
def exploded_hex:
def stream:
recurse(if . > 0 then ./16|floor else empty end) | . % 16 ;
if . == 0 then [48]
else [stream] | reverse | .[1:]
|  map(if . < 10 then 48 + . else . + 87 end)
end;```

```# Output: a stream of [decimal, hexadecimal] values
def palindromic_primes_in_base_16:
(2, (range(3; infinite; 2) | select(is_prime)))
| exploded_hex as \$hex
|select( \$hex | (. == reverse))
| [., (\$hex|implode)] ;

emit_until(.[0] >= 500; palindromic_primes_in_base_16)```
Output:
```[2,"2"]
[3,"3"]
[5,"5"]
[7,"7"]
[11,"b"]
[13,"d"]
[17,"11"]
[257,"101"]
[337,"151"]
[353,"161"]
[401,"191"]
[433,"1b1"]
[449,"1c1"]
```

## Julia

```using Primes

ispal(n, base) = begin dig = digits(n, base=base); dig == reverse(dig) end

palprimes(N, base=16) = [string(i, base=16) for i in primes(N) if ispal(i, base)]

foreach(s -> print(s, " "), palprimes(500, 16)) # 2 3 5 7 b d 11 101 151 161 191 1b1 1c1
```

## Mathematica /Wolfram Language

Giving the base 10 numbers and the base 16 numbers:

```Select[Range[499], PrimeQ[#] \[And] PalindromeQ[IntegerDigits[#, 16]] &]
BaseForm[%, 16]
```
Output:
```{2, 3, 5, 7, 11, 13, 17, 257, 337, 353, 401, 433, 449}
{Subscript[2, 16],Subscript[3, 16],Subscript[5, 16],Subscript[7, 16],Subscript[b, 16],Subscript[d, 16],Subscript[11, 16],Subscript[101, 16],Subscript[151, 16],Subscript[161, 16],Subscript[191, 16],Subscript[1b1, 16],Subscript[1c1, 16]}
```

## Nim

```import strformat, strutils

func isPalindromic(s: string): bool =
for i in 1..s.len:
if s[i-1] != s[^i]: return false
result = true

func isPrime(n: Natural): bool =
if n < 2: return false
if n mod 2 == 0: return n == 2
if n mod 3 == 0: return n == 3
var d = 5
while d * d <= n:
if n mod d == 0: return false
inc d, 2
if n mod d == 0: return false
inc d, 4
return true

var list: seq[string]
for n in 0..<500:
let h = &"{n:x}"
if h.isPalindromic and n.isPrime: list.add h

echo "Found ", list.len, " palindromic primes in base 16:"
echo list.join(" ")
```
Output:
```Found 13 palindromic primes in base 16:
2 3 5 7 b d 11 101 151 161 191 1b1 1c1```

## Perl

```    (1 x \$_ ) !~ /^(11+)\1+\$/  # test if prime
and \$h = sprintf "%x", \$_      # convert to hex
and \$h eq reverse \$h           # palindromic?
and print "\$h "                # much rejoicing
for 1..500;
```
Output:
`1 2 3 5 7 b d 11 101 151 161 191 1b1 1c1`

## Phix

```with javascript_semantics
function palindrome(string s) return s=reverse(s) end function
sequence res = filter(apply(true,sprintf,{{"%x"},get_primes_le(500)}),palindrome)
printf(1,"found %d: %s\n",{length(res),join(res)})
```
Output:
```found 13: 2 3 5 7 B D 11 101 151 161 191 1B1 1C1
```

## Quackery

See Palindromic primes#Quackery for rest of code. This is a trivial modification.

```16 base put
500 times
[ i^ isprime if
[ i^ digits palindromic if
[ i^ echo sp ] ] ]
base release```
Output:
`2 3 5 7 B D 11 101 151 161 191 1B1 1C1`

## Raku

Trivial modification of Palindromic primes task.

```say "{+\$_} matching numbers:\n{.batch(10)».fmt('%3X').join: "\n"}"
given (^500).grep: { .is-prime and .base(16) eq .base(16).flip };
```
Output:
```13 matching numbers:
2   3   5   7   B   D  11 101 151 161
191 1B1 1C1```

## REXX

```/*REXX program  finds and displays  hexadecimal palindromic primes  for all  N  <  500. */
parse arg hi cols .                              /*obtain optional argument from the CL.*/
if   hi=='' |   hi==","  then   hi= 500          /*Not specified?  Then use the default.*/
if cols=='' | cols==","  then cols=  10          /* "      "         "   "   "     "    */
call genP                                        /*build array of semaphores for primes.*/
w= 8                                             /*max width of a number in any column. */
title= ' palindromic primes in base 16 that are  < '   hi
if cols>0 then say ' index │'center(title,   1 + cols*(w+1)     )
if cols>0 then say '───────┼'center(""   ,   1 + cols*(w+1), '─')
finds= 0;                   idx= 1               /*define # of palindromic primes & idx.*/
\$=                                               /*hex palindromic primes list (so far).*/
do j=1  for hi;       if \!.j  then iterate  /*J (decimal) not prime?     Then skip.*/      /* ◄■■■■■■■■ a filter. */
x= d2x(j);  if x\==reverse(x)  then iterate  /*Hex value not palindromic?   "    "  */      /* ◄■■■■■■■■ a filter. */
finds= finds + 1                             /*bump the number of palindromic primes*/
if cols<0                      then iterate  /*Build the list  (to be shown later)? */
\$= \$  right( lowerHex(x), w)                 /*use a lowercase version of the hex #.*/
if finds//cols\==0             then iterate  /*have we populated a line of output?  */
say center(idx, 7)'│'  substr(\$, 2);   \$=    /*display what we have so far  (cols). */
idx= idx + cols                              /*bump the  index  count for the output*/
end   /*j*/

if \$\==''  then say center(idx, 7)"│"  substr(\$, 2)  /*possible display residual output.*/
if cols>0 then say '───────┴'center(""  ,   1 + cols*(w+1), '─')
say
say 'Found '         finds          title
exit 0                                           /*stick a fork in it,  we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
lowerHex: return translate( arg(1), 'abcdef', "ABCDEF") /*convert hex chars──►lowercase.*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
genP: !.= 0;  hip= max(hi, copies(9,length(hi))) /*placeholders for primes (semaphores).*/
@.1=2;  @.2=3;  @.3=5;  @.4=7;  @.5=11     /*define some low primes.              */
!.2=1;  !.3=1;  !.5=1;  !.7=1;  !.11=1     /*   "     "   "    "     flags.       */
#=5;     sq.#= @.# **2     /*number of primes so far;     prime². */
/* [↓]  generate more  primes  ≤  high.*/
do j=@.#+2  by 2  to hip                 /*find odd primes from here on.        */
parse var j '' -1 _;       if    _==5  then iterate  /*J ÷ by 5?  (right digit).*/
if j//3==0  then iterate;  if j//7==0  then iterate  /*" "  " 3?     J ÷ by 7?  */
do k=5  while sq.k<=j             /* [↓]  divide by the known odd primes.*/
if j // @.k == 0  then iterate j  /*Is  J ÷ X?  Then not prime.     ___  */
end   /*k*/                       /* [↑]  only process numbers  ≤  √ J   */
#= #+1;    @.#= j;   sq.#= j*j;   !.j= 1 /*bump # of Ps; assign next P;  P²; P# */
end          /*j*/;               return
```
output   when using the default inputs:
``` index │                       palindromic primes in base 16 that are  <  500
───────┼───────────────────────────────────────────────────────────────────────────────────────────
1   │        2        3        5        7        b        d       11      101      151      161
11   │      191      1b1      1c1
───────┴───────────────────────────────────────────────────────────────────────────────────────────

Found  13  palindromic primes in base 16 that are  <  500
```

## Ring

```load "stdlib.ring"
see "working..." + nl
see "Palindromic primes in base 16:" + nl
row = 0
limit = 500

for n = 1 to limit
hex = hex(n)
if ispalindrome(hex) and isprime(n)
see "" + upper(hex) + " "
row = row + 1
if row%5 = 0
see nl
ok
ok
next

see nl + "Found " + row + " palindromic primes in base 16" + nl
see "done..." + nl```
Output:
```working...
Palindromic primes in base 16:
2 3 5 7 B
D 11 101 151 161
191 1B1 1C1
Found 13 palindromic primes in base 16
done...
```

## Ruby

```res = Prime.each(500).filter_map do |pr|
str = pr.to_s(16)
str if str == str.reverse
end
puts res.join(", ")
```
Output:
```2, 3, 5, 7, b, d, 11, 101, 151, 161, 191, 1b1, 1c1
```

## Seed7

```\$ include "seed7_05.s7i";

const func boolean: isPrime (in integer: number) is func
result
var boolean: prime is FALSE;
local
var integer: upTo is 0;
var integer: testNum is 3;
begin
if number = 2 then
prime := TRUE;
elsif odd(number) and number > 2 then
upTo := sqrt(number);
while number rem testNum <> 0 and testNum <= upTo do
testNum +:= 2;
end while;
prime := testNum > upTo;
end if;
end func;

const proc: main is func
local
var integer: n is 0;
var string: hex is "";
begin
for n range 2 to 499 do
if isPrime(n) then
if hex = reverse(hex) then
write(hex <& " ");
end if;
end if;
end for;
end func;```
Output:
```2 3 5 7 b d 11 101 151 161 191 1b1 1c1
```

## Sidef

```func palindromic_primes(upto, base = 10) {
var list = []
for (var p = 2; p <= upto; p = p.next_palindrome(base)) {
list << p if p.is_prime
}
return list
}

var list = palindromic_primes(500, 16)

list.each {|p|
say "#{'%3s' % p}_10 = #{'%3s' % p.base(16)}_16"
}
```
Output:
```  2_10 =   2_16
3_10 =   3_16
5_10 =   5_16
7_10 =   7_16
11_10 =   b_16
13_10 =   d_16
17_10 =  11_16
257_10 = 101_16
337_10 = 151_16
353_10 = 161_16
401_10 = 191_16
433_10 = 1b1_16
449_10 = 1c1_16
```

## Wren

Library: Wren-math
Library: Wren-fmt
```import "./math" for Int
import "./fmt" for Conv, Fmt

System.print("Primes < 500 which are palindromic in base 16:")
var primes = Int.primeSieve(500)
var count = 0
for (p in primes) {
var hp = Conv.Itoa(p, 16)
if (hp == hp[-1..0]) {
Fmt.write("\$3s ", hp)
count = count + 1
if (count % 5 == 0) System.print()
}
}
System.print("\n\nFound %(count) such primes.")
```
Output:
```Primes < 500 which are palindromic in base 16:
2   3   5   7   B
D  11 101 151 161
191 1B1 1C1

Found 13 such primes.
```

## XPL0

```func IsPrime(N);        \Return 'true' if N is a prime number
int  N, I;
[if N <= 1 then return false;
for I:= 2 to sqrt(N) do
if rem(N/I) = 0 then return false;
return true;
];

func Reverse(N, Base);  \Reverse order of digits in N for given Base
int  N, Base, M;
[M:= 0;
repeat  N:= N/Base;
M:= M*Base + rem(0);
until   N=0;
return M;
];

int Count, N;
[SetHexDigits(1);
Count:= 0;
for N:= 0 to 500-1 do
if IsPrime(N) & N=Reverse(N, 16) then
[HexOut(0, N);
Count:= Count+1;
if rem(Count/10) = 0 then CrLf(0) else ChOut(0, 9\tab\);
];
CrLf(0);
IntOut(0, Count);
Text(0, " such numbers found.
");
]```
Output:
```2       3       5       7       B       D       11      101     151     161
191     1B1     1C1
13 such numbers found.
```